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An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $21.2$ 21.2, and the variance is known to be $34.81$ 34.81. How large of a sample would be required in order to estimate the mean per capita income at the 80%80% level of confidence with an error of at most $0.52$ 0.52? Round your answer up to the next integer.

Respuesta :

Answer:

We need a sample size of at least 170.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.8}{2} = 0.1[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.1 = 0.9[/tex], so [tex]z = 1.28[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population(square root of the variance) and n is the size of the sample.

How large of a sample would be required in order to estimate the mean per capita income at the 80%80% level of confidence with an error of at most $0.52$ 0.52?

A sample size of at least n, in which n is found when [tex]M = 0.52, \sigma = \sqrt{34.81} = 5.9[/tex]

So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.58 = 1.28*\frac{5.9}{\sqrt{n}}[/tex]

[tex]0.58\sqrt{n} = 1.28*5.9[/tex]

[tex]\sqrt{n} = \frac{1.28*5.9}{0.58}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.28*5.9}{0.58})^{2}[/tex]

[tex]n = 169.5[/tex]

Rounding up

We need a sample size of at least 170.

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